Left Termination of the query pattern g_in_1(a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

g(W) :- ','(eq(X, .(.(a, []), .(.(R, []), []))), ','(eq(Y, .(.(S, .(c, [])), .([], []))), ','(app_1(X, Y, Z), ','(eq(Z, .(U, V)), app_2(U, U, W))))).
app_1([], X, X).
app_1(.(X, Xs), Ys, .(X, Zs)) :- app_1(Xs, Ys, Zs).
app_2([], X, X).
app_2(.(X, Xs), Ys, .(X, Zs)) :- app_2(Xs, Ys, Zs).
eq(X, X).

Queries:

g(a).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
g_in: (f)
app_1_in: (f,f,f)
app_2_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

G_IN_A(W) → U1_A(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
G_IN_A(W) → EQ_IN_AA(X, .(.(a, []), .(.(R, []), [])))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_A(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → EQ_IN_AA(Y, .(.(S, .(c, [])), .([], [])))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_A(W, app_1_in_aaa(X, Y, Z))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → APP_1_IN_AAA(X, Y, Z)
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U6_AAA(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)
U3_A(W, app_1_out_aaa(X, Y, Z)) → U4_A(W, eq_in_aa(Z, .(U, V)))
U3_A(W, app_1_out_aaa(X, Y, Z)) → EQ_IN_AA(Z, .(U, V))
U4_A(W, eq_out_aa(Z, .(U, V))) → U5_A(W, app_2_in_aaa(U, U, W))
U4_A(W, eq_out_aa(Z, .(U, V))) → APP_2_IN_AAA(U, U, W)
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U7_AAA(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA
U1_A(x1, x2)  =  U1_A(x2)
G_IN_A(x1)  =  G_IN_A
U7_AAA(x1, x2, x3, x4, x5)  =  U7_AAA(x5)
EQ_IN_AA(x1, x2)  =  EQ_IN_AA
U4_A(x1, x2)  =  U4_A(x2)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2, x3)  =  U2_A(x3)
U6_AAA(x1, x2, x3, x4, x5)  =  U6_AAA(x5)
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA
U5_A(x1, x2)  =  U5_A(x2)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

G_IN_A(W) → U1_A(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
G_IN_A(W) → EQ_IN_AA(X, .(.(a, []), .(.(R, []), [])))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_A(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → EQ_IN_AA(Y, .(.(S, .(c, [])), .([], [])))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_A(W, app_1_in_aaa(X, Y, Z))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → APP_1_IN_AAA(X, Y, Z)
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U6_AAA(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)
U3_A(W, app_1_out_aaa(X, Y, Z)) → U4_A(W, eq_in_aa(Z, .(U, V)))
U3_A(W, app_1_out_aaa(X, Y, Z)) → EQ_IN_AA(Z, .(U, V))
U4_A(W, eq_out_aa(Z, .(U, V))) → U5_A(W, app_2_in_aaa(U, U, W))
U4_A(W, eq_out_aa(Z, .(U, V))) → APP_2_IN_AAA(U, U, W)
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U7_AAA(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA
U1_A(x1, x2)  =  U1_A(x2)
G_IN_A(x1)  =  G_IN_A
U7_AAA(x1, x2, x3, x4, x5)  =  U7_AAA(x5)
EQ_IN_AA(x1, x2)  =  EQ_IN_AA
U4_A(x1, x2)  =  U4_A(x2)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2, x3)  =  U2_A(x3)
U6_AAA(x1, x2, x3, x4, x5)  =  U6_AAA(x5)
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA
U5_A(x1, x2)  =  U5_A(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 12 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

Q DP problem:
The TRS P consists of the following rules:

APP_2_IN_AAA → APP_2_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_2_IN_AAA → APP_2_IN_AAA

The TRS R consists of the following rules:none


s = APP_2_IN_AAA evaluates to t =APP_2_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_2_IN_AAA to APP_2_IN_AAA.





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

Pi DP problem:
The TRS P consists of the following rules:

APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

Q DP problem:
The TRS P consists of the following rules:

APP_1_IN_AAA → APP_1_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_1_IN_AAA → APP_1_IN_AAA

The TRS R consists of the following rules:none


s = APP_1_IN_AAA evaluates to t =APP_1_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_1_IN_AAA to APP_1_IN_AAA.




We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
g_in: (f)
app_1_in: (f,f,f)
app_2_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

G_IN_A(W) → U1_A(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
G_IN_A(W) → EQ_IN_AA(X, .(.(a, []), .(.(R, []), [])))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_A(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → EQ_IN_AA(Y, .(.(S, .(c, [])), .([], [])))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_A(W, app_1_in_aaa(X, Y, Z))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → APP_1_IN_AAA(X, Y, Z)
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U6_AAA(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)
U3_A(W, app_1_out_aaa(X, Y, Z)) → U4_A(W, eq_in_aa(Z, .(U, V)))
U3_A(W, app_1_out_aaa(X, Y, Z)) → EQ_IN_AA(Z, .(U, V))
U4_A(W, eq_out_aa(Z, .(U, V))) → U5_A(W, app_2_in_aaa(U, U, W))
U4_A(W, eq_out_aa(Z, .(U, V))) → APP_2_IN_AAA(U, U, W)
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U7_AAA(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA
U1_A(x1, x2)  =  U1_A(x2)
G_IN_A(x1)  =  G_IN_A
U7_AAA(x1, x2, x3, x4, x5)  =  U7_AAA(x5)
EQ_IN_AA(x1, x2)  =  EQ_IN_AA
U4_A(x1, x2)  =  U4_A(x2)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2, x3)  =  U2_A(x3)
U6_AAA(x1, x2, x3, x4, x5)  =  U6_AAA(x5)
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA
U5_A(x1, x2)  =  U5_A(x2)

We have to consider all (P,R,Pi)-chains

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

G_IN_A(W) → U1_A(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
G_IN_A(W) → EQ_IN_AA(X, .(.(a, []), .(.(R, []), [])))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_A(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U1_A(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → EQ_IN_AA(Y, .(.(S, .(c, [])), .([], [])))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_A(W, app_1_in_aaa(X, Y, Z))
U2_A(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → APP_1_IN_AAA(X, Y, Z)
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U6_AAA(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)
U3_A(W, app_1_out_aaa(X, Y, Z)) → U4_A(W, eq_in_aa(Z, .(U, V)))
U3_A(W, app_1_out_aaa(X, Y, Z)) → EQ_IN_AA(Z, .(U, V))
U4_A(W, eq_out_aa(Z, .(U, V))) → U5_A(W, app_2_in_aaa(U, U, W))
U4_A(W, eq_out_aa(Z, .(U, V))) → APP_2_IN_AAA(U, U, W)
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U7_AAA(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA
U1_A(x1, x2)  =  U1_A(x2)
G_IN_A(x1)  =  G_IN_A
U7_AAA(x1, x2, x3, x4, x5)  =  U7_AAA(x5)
EQ_IN_AA(x1, x2)  =  EQ_IN_AA
U4_A(x1, x2)  =  U4_A(x2)
U3_A(x1, x2)  =  U3_A(x2)
U2_A(x1, x2, x3)  =  U2_A(x3)
U6_AAA(x1, x2, x3, x4, x5)  =  U6_AAA(x5)
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA
U5_A(x1, x2)  =  U5_A(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 12 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

Pi DP problem:
The TRS P consists of the following rules:

APP_2_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_2_IN_AAA(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
APP_2_IN_AAA(x1, x2, x3)  =  APP_2_IN_AAA

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP

Q DP problem:
The TRS P consists of the following rules:

APP_2_IN_AAA → APP_2_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_2_IN_AAA → APP_2_IN_AAA

The TRS R consists of the following rules:none


s = APP_2_IN_AAA evaluates to t =APP_2_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_2_IN_AAA to APP_2_IN_AAA.





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

g_in_a(W) → U1_a(W, eq_in_aa(X, .(.(a, []), .(.(R, []), []))))
eq_in_aa(X, X) → eq_out_aa(X, X)
U1_a(W, eq_out_aa(X, .(.(a, []), .(.(R, []), [])))) → U2_a(W, X, eq_in_aa(Y, .(.(S, .(c, [])), .([], []))))
U2_a(W, X, eq_out_aa(Y, .(.(S, .(c, [])), .([], [])))) → U3_a(W, app_1_in_aaa(X, Y, Z))
app_1_in_aaa([], X, X) → app_1_out_aaa([], X, X)
app_1_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U6_aaa(X, Xs, Ys, Zs, app_1_in_aaa(Xs, Ys, Zs))
U6_aaa(X, Xs, Ys, Zs, app_1_out_aaa(Xs, Ys, Zs)) → app_1_out_aaa(.(X, Xs), Ys, .(X, Zs))
U3_a(W, app_1_out_aaa(X, Y, Z)) → U4_a(W, eq_in_aa(Z, .(U, V)))
U4_a(W, eq_out_aa(Z, .(U, V))) → U5_a(W, app_2_in_aaa(U, U, W))
app_2_in_aaa([], X, X) → app_2_out_aaa([], X, X)
app_2_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U7_aaa(X, Xs, Ys, Zs, app_2_in_aaa(Xs, Ys, Zs))
U7_aaa(X, Xs, Ys, Zs, app_2_out_aaa(Xs, Ys, Zs)) → app_2_out_aaa(.(X, Xs), Ys, .(X, Zs))
U5_a(W, app_2_out_aaa(U, U, W)) → g_out_a(W)

The argument filtering Pi contains the following mapping:
g_in_a(x1)  =  g_in_a
U1_a(x1, x2)  =  U1_a(x2)
eq_in_aa(x1, x2)  =  eq_in_aa
eq_out_aa(x1, x2)  =  eq_out_aa
U2_a(x1, x2, x3)  =  U2_a(x3)
U3_a(x1, x2)  =  U3_a(x2)
app_1_in_aaa(x1, x2, x3)  =  app_1_in_aaa
app_1_out_aaa(x1, x2, x3)  =  app_1_out_aaa
U6_aaa(x1, x2, x3, x4, x5)  =  U6_aaa(x5)
U4_a(x1, x2)  =  U4_a(x2)
U5_a(x1, x2)  =  U5_a(x2)
app_2_in_aaa(x1, x2, x3)  =  app_2_in_aaa
app_2_out_aaa(x1, x2, x3)  =  app_2_out_aaa
U7_aaa(x1, x2, x3, x4, x5)  =  U7_aaa(x5)
g_out_a(x1)  =  g_out_a
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

APP_1_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_1_IN_AAA(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
APP_1_IN_AAA(x1, x2, x3)  =  APP_1_IN_AAA

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof

Q DP problem:
The TRS P consists of the following rules:

APP_1_IN_AAA → APP_1_IN_AAA

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_1_IN_AAA → APP_1_IN_AAA

The TRS R consists of the following rules:none


s = APP_1_IN_AAA evaluates to t =APP_1_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_1_IN_AAA to APP_1_IN_AAA.